# How do you factor 12n^3+15n^2+4n+5?

$12 {n}^{3} + 15 {n}^{2} + 4 n + 5$
By inspection the first two terms are $3 {n}^{2}$ times the last two terms
$12 {n}^{3} + 15 {n}^{2} + 4 n + 5$
$= \left(3 {n}^{2}\right) \left(4 n + 5\right) + \left(1\right) \left(4 n + 5\right)$
$= \left(3 {n}^{2} + 1\right) \left(4 n + 5\right)$